Originally Posted by
Captcha It's not $\displaystyle \left<x\right>^2$ I want to find. It's $\displaystyle \left<x^2\right>
$.
Basically I'm trying to solve for $\displaystyle x$ the equation:
$\displaystyle 2.61 = \sqrt{\left<x^2\right>-\left<x\right>^2}$ to find the value of $\displaystyle x$, where $\displaystyle 2.61$ is the $\displaystyle \text{standard deviation}$ of the data $\displaystyle 8,7, 4, 10, 6, x$.
I know that $\displaystyle \left<x\right> = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right)$.
$\displaystyle \Rightarrow \left<x^2\right> = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2$.
So $\displaystyle 2.61 = \sqrt{\left<x^2\right>-\frac{1}{36}\left(35+x\right)^2}$
But I've no idea how to get $\displaystyle \left<x^2\right> $.